Answer:
approximately 0.0969
Explanation:
Well this is quite tricky but, to determine if the normal approximation is appropriate, we need to check if the conditions for using the normal distribution are met. The conditions are:
The sample size is large enough, typically n ≥ 30.
The sample is drawn at random.
The observations are independent of each other.
The population distribution is approximately normal or the sample size is large enough that the central limit theorem applies.
So, in this case, we are given that 60 cars are selected at random, so the first condition is satisfied.
The second and third conditions are assumed since we are not given any information to suggest otherwise. The fourth condition is more difficult to assess, as we do not know the population distribution of cars using the HOV lane. However, we can use the rule of thumb that if the sample size is large enough (n ≥ 30), the central limit theorem applies and we can use the normal distribution to model the sample proportion.
So, we can conclude that the normal approximation is appropriate for this situation.
Let p be the proportion of cars that have less than 2 people in them. We are given that approximately 20% of the time, a driver on their own uses the HOV lane, which means that p = 0.2.
We want to find the probability that exactly 10 cars have less than 2 people in them. Let X be the number of cars with less than 2 people in them, then X follows a binomial distribution with parameters n = 60 and p = 0.2. We can use the normal approximation to the binomial distribution, which is:
X - N(np, np(1-p)
So, the mean and standard deviation of X are:
mean = np = 60 × 0.2 = 12
standard deviation = sqrt(np(1-p)) = sqrt(60 × 0.2 × 0.8) = 2.19
To find the probability that exactly 10 cars have less than 2 people in them, we need to find the probability that X = 10:
P(X = 10) = P(9.5 ≤ X ≤ 10.5) (using continuity correction)
We can standardize the distribution to a standard normal distribution:
z = (X - mean) / standard deviation = (10 - 12) / 2.19 = -0.91
P(9.5 ≤ X ≤ 10.5) = P(-0.91 ≤ z ≤ -0.46) (using the standard normal distribution table or calculator)
Looking up the standard normal distribution table or using a calculator, we find that P(-0.91 ≤ Z ≤ -0.46) = 0.0969.>>
So therefore, the probability that exactly 10 cars have less than 2 people in them is approximately 0.0969..